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If = 2 6^ + 126^ 3 , then which of the following quadratic equations has as one of its roots?

Options

  1. A4x^2 + 2x - 1 = 0
  2. B4x^2 - 2x - 1 = 0
  3. C16x^2 - 8x - 1 = 0
  4. D4x^2 - x - 1 = 0

Correct answer

B. 4x^2 - 2x - 1 = 0

Step-by-step solution

We are given = 2 6^ + 126^ 3 . Split the term 2 6^ to apply the sum-to-product formula: = 6^ + 6^ + 126^ 3 Using the formula C + D = 2 ( C+D 2 ) ( C-D 2 ) : 6^ + 126^ = 2 66^ 60^ = 2 66^ ( 1 2 ) = 66^ Substitute this back into the expression for : = 6^ + 66^ 3 Apply the sum-to-product formula again: 6^ + 66^ = 2 36^ 30^ = 2 36^ ( 3 2 ) = 3 36^ Thus, = 3 36^ 3 = 36^ . We know that 36^ = 5 + 1 4 . Therefore, = 5 + 1 4 4 - 1 = 5 Squaring both sides: (4 - 1)^2 = 5 16 ^2 - 8 + 1 = 5 16 ^2 - 8 - 4 = 0 4 ^2 - 2 - 1 = 0 So

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